Ex. 2.

Find the square root of 12-140.

Here a=12

... a'-b=144-140=4, and a-b=2.

.. a-b=961+720=1681, and a-b

=41.

Hence √§a+§√ a−b+ √ § a − § √ a ® − b =

Ex. 4. Find the square root of 7+4√3. ANSW.2+/3.

Ex. 5.

Ex. 6.

7-210 ... √5-√2.

.18-10-7... 5-√-7.

177

CHAP. IX.

ON

THE INVESTIGATION OF THE BINOMIAL THEOREM;

AND ON THE

EXPANSION AND REVERSION OF SERIES.

PREVIOUS

REVIOUS to the investigation of the Binomial Theorem, it is necessary to explain the manner of finding the number of permutations or combinations which can take place amongst any set of quantities.

XLI.

On Permutations and Combinations.

137. By Permutations are meant the number of changes which any quantities a, b, c, d, &c. may undergo with respect to their order, when taken two and two together, three and three, &c. &c. Thus ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc, are the different permutations of the four quantities a, b, c, d, when taken two and two together; abc, acb, bac, bca, cab, cba, of the three quantities a, b, c, when taken three and three together; &c. &c.

138. By Combinations are meant the number of collections which may be formed out of the quantity a, b, c, d, e, &c. taken two and two together, three and three together, &c. &c. without

2 A

without having regard to the order in which the quantities are arranged in each collection. Thus ab, ac, ad, bc, bd, cd, are the combinations which can be formed out of the four quantities a, b, c, d, taken two and two together; abc, abd, acd, bcd, the combinations which may be formed out of the same quantities, when taken three and three together; &c. &c.

139. Let there be n quantities, a, b, c, d, e, &c., then, by Art. 137, it appears that there will be (n-1) permutations in which a stands first; for the same reason there will be (n-1) permutations in which stands first; and so of c, d, e, &c. Hence there will be n times (n-1) permutations of the form ab, ac, ad, ae, &c.; ba, bc, bd, be, &c.; ca, cb, cd, ce, &c.; i.e. "the number of permutations of n things taken two and two is n(n − 1).”

140. If these n quantities be taken three and three together, then there will be n(n − 1) (n − 2) permutations. For if (n−1) be substituted for n in the last article, then the number of permutations of n-1 things taken two and two together will be (n-1)(n-2); hence the number of permutations of b, c, d, e, &c. taken two and two together,are (n − 1)(n— 2), and consequently there are (n−1)(n−2) permutations of the quantities a, b, c, d, e, &c. taken three and three together, in which a may stand first; for the same reason there are (n−1)(n-2) permutations in which may stand first; and so of c, d, e, &c. The numbers of the permutations of this kind will therefore amount to n(n − 1) (n − 2).

b

141. In general therefore it appears, that if the number of quantities ben, and they are taken m and m together, the number of permutations will be n(n-1) (n-2) &c........ (n-m+1); and if m=n, i.e. if the permutations respect all the quantities at once, then (since m―n=0) the number of them will be n(n - 1) (n − 2) &c.....2.1. Thus, the number of

permu

permutations which might be formed from the letters composing the word "virtue” are 6 × 5 × 4 × 3 × 2 × 1=720.

142. But if the same letter should occur any number of times, then it is evident that we must divide the whole number of permutations, by the number of times the permutations are increased by having different letters instead of the repetition of the same letter. Thus if the same letter should occur twice, then we must divide by 2 x 1; if it should occur thrice, we must divide by 3×2× 1; if p times, by 1.2.3...p; and so for any other letter which may occur more than once. Hence the general expression for the number of permutations of n things, of which there are p of one kind; r of another; n(n-1) (n-2)n-3....2.1

g of another; &c. &c. is

Thus

1.2.3..px 1.2.3..r x 1.2.3..q the permutations which may be formed by the letters composing the word "easiness" (since s occurs thrice, e twice) 8.7.6.5.4.3.2.1

are

1.2.3. X 1.2

3360.

143. From the expression (in Art. 141) for finding the number of permutations of n things taker m and m together, we immediately deduce the theorem for finding the number of combinations of n things taken in the same manner. For the permutations of n things taken two and two together being n(n-1), and as each combination admits of as many permutations as may be made by two things (which is 2 x 1), the number of combinations must be equal to the number of permutations divided by 2; i.e. the number of combinations of n(n-1) n things taken two and two together is

2

For the

same reason, the combinations of n things, taken three and three n(n−1) (n—2); and in general, the

together, must be equal to

1.2.3

combinations of n things taken m and m together must be equal

x + a x+b

XLII.

On the Investigation of the Binomial Theorem.

144, In Chap. III. it was shewn, that, by raising the successive powers of (a+b) in the ordinary way of multiplication, the nth power of a+b might be exhibited under the form a"+na"-b n (n − 1) ausb2 + n (n − 1) (n−2) ab3+&c. For the inves

+

1.2

1.2.3

tigation of this Theorem, let us suppose the quantities, (x+a),
(x+b), (x+c), (x+d), &c. multiplied together, as in the
following Table,

x2+ax

+ bx+ab

x2+(a+b)x+ab=(x+a) (x + b)

x + c

x2+(a+b)x2+ab.x

+c.x2+(ac+bc) x + a b c

x13+(a+b+c) x2 +(ab+ac+bc)x+abc=(x+a)(x+1)(x+c) x+d

.18* + (a + b + c) x3 +(ab+ac+bc)x+abc.x

+d.x2+(ad+bd+cd) x2+(abd+acd+bcd)x+abcd x*+(a+b+c+d)x3 + (ab+ac+ad+be+bd+cd)x2+(abc+abd+acd+be +abcd=(x+a)(x+b)(x+c)(x+d)

145. By the mere inspection of this Table, it appears that if there had been n quantities a, b, c, d, e, &c. then the first term of the product would be x"; the second P-1, where P=(a+b+c+d+e), &c.; the third Qa, where

Q=